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Ostrovityanka [42]
3 years ago
5

Please answer this correctly

Mathematics
1 answer:
Readme [11.4K]3 years ago
8 0

Answer:

2MIN

Step-by-step explanation:

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Which of the following equations could represebt the point in the table select all the following
r-ruslan [8.4K]

Answer:

what are the equations

Step-by-step explanation:

you didnt show the equations

6 0
3 years ago
W = A/L<br> (solve for A)<br> A = W +L<br> A = W-L<br> A = W/L<br> A = LW
kipiarov [429]
A=wl, so the answer is A :)
6 0
3 years ago
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What is the opposite of -41
Step2247 [10]
The opposite of -41 is positive 41. This is also known as the absolute value of -41
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3 years ago
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In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
3 years ago
You have a 1-gallon paint can in the shape of a cylinder. One gallon is 231 cubic inches. The radius of the can is 3 inches. Wha
aleksandr82 [10.1K]

Answer:

The approximate height of the paint can is 8.2 in

Step-by-step explanation:

we know that

The volume of the cylinder ( can of paint) is equal to

V=\pi r^{2} h

we have

V=231\ in^{3}

r=3\ in

\pi=3.14

Substitute the values and solve for h

231=(3.14)(3)^{2} h

h=231/(3.14*9)=8.2 in

3 0
3 years ago
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